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Finding Acceleration

  1. Sep 3, 2007 #1
    1. The problem statement, all variables and given/known data
    Two blocks, each of mass m, are connected by a string of constant length 4h. Block A is placed on top of a table 3h from the side and Block B hangs over the edge. The tabletop is 2h above the floor. At time t=0, Block B is released from rest at a distance 1h. Use variables h, m, and g to answer the following: Determine the acceleration of block B as it descends.

    2. Relevant equations
    [tex] a= \Sigma F/m [/tex]
    [tex] \Sigma F_x= (m)(a_x) [/tex]
    [tex] \Sigma F_y= (m)(a_y) [/tex]

    3. The attempt at a solution
    I drew a free-body sketch of Block A and Block B and tried solving it by [tex] \Sigma F_x= 0 + T= (m)(a) [/tex] and [tex] \Sigma F_y= T - W_2= (m)(-a) [/tex]. This is where I get stuck because I try and plug T from the first equation into the second one and it doesn't end up right (I get a=a/g or g=1).
  2. jcsd
  3. Sep 3, 2007 #2
    You have the forces in the x-direction for block A incorrect. What is the normal force of block A and what is it causing it to do? What is the tension trying to do to that same block? If there is an imbalance between the normal force, which is trying to keep it from accelerating, and the tension which is trying to pull it toward the right, what happens? what about acceleration on the two blocks? are they going to be different or the same?
  4. Sep 3, 2007 #3
    Okay, for [tex] \Sigma F_x= -Wsin(180) + T. [/tex] I think that is right because [tex] F_N [/tex] is a force that the table exerts up in the +y direction on Block A while W is points in the -y direction. The tension is moving Block A in the +x direction. Acceleration is the same for both blocks because the rope length remains at a constant of 4h.
    Last edited: Sep 3, 2007
  5. Sep 3, 2007 #4


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    Your original post solution looks right... except you made an algebra mistake... a does not come out to g.
  6. Sep 3, 2007 #5
    [tex] \Sigma F_x = 0 + T = (m)(a) [/tex] so T= (m)(a). I plug that into [tex] \Sigma F_y= T - W_2= (m)(-a) [/tex] and come out with [tex] \Sigma F_y= (m)(a) - W_2= (m)(-a) [/tex] and substitute W for (m)(g). Now the equation looks like (m)(a) -(m)(g). (m)(a)= (m)(g) then divide by m and a=g, that's what I keep getting.
  7. Sep 3, 2007 #6


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    ma - mg = -ma

    2ma = mg

    a = g/2
  8. Sep 3, 2007 #7
    Okay, I see that I was supposed to add (m)(-a)= [tex] \Sigma F_y [/tex] Thanks for the help.
    Last edited: Sep 3, 2007
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